Energy-stress tensor integration proof (from schutz ch.4)

1. The problem statement, all variables and given/known data
Use the identity [tex]T^{\mu \nu}_{ ,\nu} = 0[/tex] to prove the following results for a bounded system (ie a system for which [tex]T^{\mu \nu} = 0[/tex]
outside a bounded region of space),

[tex]\frac{\partial}{\partial t}\int T^{0\alpha}d^{3}x = 0[/tex]

2. Relevant equations

3. The attempt at a solution

The integral obviously gives 4 equations (one for each [tex]\alpha[/tex]) which must all be = 0.
I tried just working on the first and passing the partial derivative into the integral and writing [tex]\frac{\partial}{\partial t} T^{0 0} = -\left(\frac{\partial}{\partial x} T^{0 1} +\frac{\partial}{\partial y} T^{0 2} + \frac{\partial}{\partial z} T^{0 3} \right)[/tex]

This gives
[tex]- \int T^{0 i}_{,i}d^{3}x[/tex]

which doeesnt really seem to be getting me anywhere - also the same reasoning wouldn't work for the other equations because they must all be partially differentiated wrt t and this is only relevant for [tex]T^{0 0}[/tex] in [tex]T^{\mu \nu}_{ ,\nu} = 0[/tex]

Another direction I thought of was to use Gauss' law but then there is no outward normal one-form and so maybe not....

There are another two parts to this question but I thought that if I had an idea of how to do the first part I could figure the others out by myself.

Since the system is bounded, why not use [itex]T^{0 \alpha}=T^{\alpha 0}[/itex]? Then [tex] \frac{\partial}{\partial t}T^{0 \alpha}=\frac{\partial}{\partial t}T^{\alpha 0}=T^{\alpha 0}_{,0}[/tex]....but what is this last quantity if [itex]T^{\mu \nu}_{ ,\nu} = 0[/itex]? ;0)

Well, what do the components [tex]T^{\alpha i}[/tex] represent physically?

The [itex]\alpha[/itex] momentum per second crossing a unit area of surface of constant [itex]x^{i}[/itex]

So does the " constant [itex]x^{i}[/itex] " mean that [itex]x_{1}=x_{2}[/itex], [itex]y_{1}=y_{2}[/itex] and [itex]z_{1}=z_{2}[/itex] and therefore the integrands in the above integrals are all zero and so the integrals are also zero?

I feel a bit like I'm clutching at straws here...
Is that right or am I just making stuff up? :D

But the second proof made me realise that this doesnt really make sense.... why should [itex]x_{1}=x_{2}[/itex]? these values are unrelated to [itex]T^{\alpha 1}[/itex]
What would make more sense (and satisfy both proofs) is that [itex]T^{\alpha 1} (x_{1}) = T^{\alpha 1}(x_{2}) = 0[/itex]
Then both
[tex] \left[T^{0k}x^{i}x^{j} \right] ^{x^{k}_2}_{x^{k}_1} = 0[/tex] (satisfying the second proof)
and
[tex]T^{\alpha 1} (x_{1}) - T^{\alpha 1} (x_{2}) = 0[/tex] (satisfying the first)

But I'm not sure why [itex]T^{\alpha 1} (x_{1}) = 0[/itex].
My logic says this is incorrect - if we have a constant x surface, then [itex]T^{\alpha 1}[/itex] doesn't depend on x and so [itex]T^{\alpha 1} (x_{1}) = T^{\alpha 1}(x_{2}) = T^{\alpha 1}[/itex] but not necessarily 0. So I have a problem with my second proof.